Q 12-03-246JEE MainJEE Main 2019 (10 Jan, Shift 1)Medium
A potentiometer wire $AB$ having length $L$ and resistance $12r$ is joined to a cell $D$ of emf $\varepsilon$ and internal resistance $r$. A cell $C$ having emf $\varepsilon/2$ and internal resistance $3r$ is connected. The length $AJ$, at which the galvanometer, as shown in the figure, shows no deflection is
Answer: (D) $\dfrac{13}{24}L$
Current in the potentiometer wire: $I = \dfrac{\varepsilon}{12r + r} = \dfrac{\varepsilon}{13r}$.
At balance no current flows through $C$, so its internal resistance does not matter. The potential drop across $AJ = l$ equals $\varepsilon/2$:
$$\frac{\varepsilon}{13r}\cdot12r\cdot\frac lL = \frac\varepsilon2 \;\Rightarrow\; l = \frac{13}{24}L$$
Solution by Sreeraj P, M.Sc Physics